2020 AMC 10A Problem 9

Let $x$ be the number of adults, which equals the number of children. Each adult takes $\frac{1}{7}$ of a section and each child $\frac{1}{11}$, so filling $N$ sections exactly means
$$\frac{x}{7} + \frac{x}{11} = N \quad\Longrightarrow\quad 18x = 77N.$$
The left side is a multiple of $18$, so $77N$ must be as well. Because $77 = 7 \cdot 11$ shares no factor with $18 = 2 \cdot 3^2$, the factor $18$ must divide $N$ itself. The least positive such $N$ is $18$, achieved with $x = 77$ adults and $77$ children.

The answer is $\boxed{\textbf{(B)}\ 18}$.

Translating "occupy all the bench space" into an equation about fractions of a section produces a linear relation with integer unknowns. The final step is pure divisibility: when $a \mid bc$ and $\gcd(a, b) = 1$, then $a \mid c$. The word-problem wrapper hides a coprimality argument.

Measure a section as $77$ units, so an adult uses $11$ units and a child $7$. Equal numbers $x$ of each use $18x$ units, which must equal $77N$; the same divisibility argument gives $N = 18$.

• Answering 77 = 7 x 11 by lcm instinct, or 9 by halving without checking that 77N/18 must be an integer.
• Setting $7N = 11N$ or otherwise assuming the adults and children each fill whole sections.